Finding Volumes Chapter 6.2 February 22, 2007

In General: Vertical Cut:Horizontal Cut:

Find the area of the region bounded by Bounds? Right Function? Left Function? Area?

Volume & Definite Integrals We used definite integrals to find areas by slicing the region and adding up the areas of the slices. We will use definite integrals to compute volume in a similar way, by slicing the solid and adding up the volumes of the slices. For Example………………

Blobs in Space Volume of a blob: Cross sectional area at height h: A(h) Volume =

Volumes: We will be given a “boundary” for the base of the shape which will be used to find a length. We will use that length to find the area of a figure generated from the slice. The dy or dx will be used to represent the thickness. The volumes from the slices will be added together to get the total volume of the figure.

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square. Bounds: Length? Area? Volume?

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a circle with diameter in the plane. Length? Area? Volume?

Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles. Length? Area? Volume? Bounds? [0,1]

The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve. Bounds? Area? Length? Volume?

Area of an Equilateral Triangle? Area = (1/2)b*h

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane. Length? Area? Bounds? Volume?

Area of Square whose diagonal is in the plane?

Solids of Revolution We start with a known planar shape and rotate that shape about a line resulting in a three dimensional shape known as a solid of revolution. When this solid of revolution takes on a non-regular shape, we can use integration to compute the volume. For example……

The Bell! Volume: Area:

Find the volume of the solid generated by revolving the region defined by, x = 3 and the x-axis about the x-axis. Bounds? Length? (radius) Area? Volume?

Solids of Revolution For solids of revolution, cross sections are circles, so we can use the formula Usually, the only difficult part is determining r(h). A good sketch is a big help. Volume =

Aside: Sketching Revolutions 1.Sketch the curve; determine the region. 2.Sketch the reflection over the axis. 3.Sketch in a few “revolution” lines.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. x = sin(y)

Example Revolve the region under the curve y = 3e –x, for 0 ≤ x ≤ 1, about the x axis.

Remember for this Method: Slices are perpendicular to the axis of rotation. Radius is a function of position on that axis. Therefore rotating about x axis gives an integral in x; rotating about y gives an integral in y.

Example Rotate y = x 2, from x = 0 to x = 4, about the x-axis. Find r(x): r(x) =

Find the volume of the solid generated by revolving the region defined by, on the interval [1,2] about the x-axis. Bounds? Length? (radius) Area? Volume?

Find the volume of the solid generated by revolving the region defined by, y = 8, and x = 0 about the y-axis. Bounds? Length? Area? Volume?

Find the volume of the solid generated by revolving the region defined by, and y = 1, about the line y = 1 Bounds? Length? Area? Volume?

*Find the volume of the solid generated by revolving the region defined by, and y = 1, about the x-axis. Bounds? Outside Radius? Inside Radius? Area? Volume?

Find the volume of the solid generated by revolving the region defined by, and y = 1, about the line y=-1. Bounds? Outside Radius? Inside Radius? Area? Volume?

Solids of Revolution For solids of revolution, cross sections are circles, If there is a gap between the function and the axis of rotation, we have a washer and use: If there is NO gap, we have a disk and use: Volume =